Question

A man running at a speed $5 \,m/s$ is viewed in the side view mirror of radius of curvature $R = 2\, m$ of a stationary car. Calculate the speed of image when the man is at a distance of $9 \,m$ from the mirror

• A. $0.3 \,m/s$
• B. $0.2 \,m/s$
• C. $0.1\, m/s$
• D. $0.05 \,m/s$

Answer 1

Answer: (D)

$0.05 \,m/s$

Answer 2

A convex mirror is used for this purpose
According to mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Differentiate w.r.t. time $t$
$\frac{d}{dt} \left[\frac{1}{V}\right]+\frac{d}{dt} \left[\frac{1}{u}\right]=\frac{d}{dt} \left[\frac{1}{f}\right]$
$-\frac{1}{v^{2}} \frac{dv}{dt}-\frac{1}{u^{2}} \frac{du}{dt}=0$
$\frac{dv}{dt}=-\frac{v^{2}}{u^{2}} \frac{du}{dt}$
Also to find $v$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=1-\left(\frac{1}{-9}\right)$
$\frac{1}{v}=\frac{9+1}{9}=\frac{10}{9}$
$\Rightarrow v=\frac{9}{10}$
$\therefore \frac{dv}{dt}=-\left(\frac{9}{10}\right)\times\frac{1}{\left(-9\right)^{2}}\times\left(5\right)$
$=0.05 m/ s$